3.13.99 \(\int \frac {(-1+x^4)^{3/4} (4+x^4)}{x^8 (-4+x^4)} \, dx\)

Optimal. Leaf size=94 \[ -\frac {3^{3/4} \tan ^{-1}\left (\frac {\sqrt [4]{3} x}{\sqrt {2} \sqrt [4]{x^4-1}}\right )}{8 \sqrt {2}}-\frac {3^{3/4} \tanh ^{-1}\left (\frac {\sqrt [4]{3} x}{\sqrt {2} \sqrt [4]{x^4-1}}\right )}{8 \sqrt {2}}+\frac {\left (x^4-1\right )^{3/4} \left (x^4+6\right )}{42 x^7} \]

________________________________________________________________________________________

Rubi [A]  time = 0.12, antiderivative size = 105, normalized size of antiderivative = 1.12, number of steps used = 7, number of rules used = 7, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.280, Rules used = {580, 583, 12, 377, 212, 206, 203} \begin {gather*} -\frac {3^{3/4} \tan ^{-1}\left (\frac {\sqrt [4]{3} x}{\sqrt {2} \sqrt [4]{x^4-1}}\right )}{8 \sqrt {2}}-\frac {3^{3/4} \tanh ^{-1}\left (\frac {\sqrt [4]{3} x}{\sqrt {2} \sqrt [4]{x^4-1}}\right )}{8 \sqrt {2}}+\frac {\left (x^4-1\right )^{3/4}}{7 x^7}+\frac {\left (x^4-1\right )^{3/4}}{42 x^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[((-1 + x^4)^(3/4)*(4 + x^4))/(x^8*(-4 + x^4)),x]

[Out]

(-1 + x^4)^(3/4)/(7*x^7) + (-1 + x^4)^(3/4)/(42*x^3) - (3^(3/4)*ArcTan[(3^(1/4)*x)/(Sqrt[2]*(-1 + x^4)^(1/4))]
)/(8*Sqrt[2]) - (3^(3/4)*ArcTanh[(3^(1/4)*x)/(Sqrt[2]*(-1 + x^4)^(1/4))])/(8*Sqrt[2])

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 203

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTan[(Rt[b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 212

Int[((a_) + (b_.)*(x_)^4)^(-1), x_Symbol] :> With[{r = Numerator[Rt[-(a/b), 2]], s = Denominator[Rt[-(a/b), 2]
]}, Dist[r/(2*a), Int[1/(r - s*x^2), x], x] + Dist[r/(2*a), Int[1/(r + s*x^2), x], x]] /; FreeQ[{a, b}, x] &&
!GtQ[a/b, 0]

Rule 377

Int[((a_) + (b_.)*(x_)^(n_))^(p_)/((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Subst[Int[1/(c - (b*c - a*d)*x^n), x]
, x, x/(a + b*x^n)^(1/n)] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && EqQ[n*p + 1, 0] && IntegerQ[n]

Rule 580

Int[((g_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.)*((e_) + (f_.)*(x_)^(n_)),
x_Symbol] :> Simp[(e*(g*x)^(m + 1)*(a + b*x^n)^(p + 1)*(c + d*x^n)^q)/(a*g*(m + 1)), x] - Dist[1/(a*g^n*(m + 1
)), Int[(g*x)^(m + n)*(a + b*x^n)^p*(c + d*x^n)^(q - 1)*Simp[c*(b*e - a*f)*(m + 1) + e*n*(b*c*(p + 1) + a*d*q)
+ d*((b*e - a*f)*(m + 1) + b*e*n*(p + q + 1))*x^n, x], x], x] /; FreeQ[{a, b, c, d, e, f, g, p}, x] && IGtQ[n
, 0] && GtQ[q, 0] && LtQ[m, -1] &&  !(EqQ[q, 1] && SimplerQ[e + f*x^n, c + d*x^n])

Rule 583

Int[((g_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.)*((e_) + (f_.)*(x_)^(n_)),
x_Symbol] :> Simp[(e*(g*x)^(m + 1)*(a + b*x^n)^(p + 1)*(c + d*x^n)^(q + 1))/(a*c*g*(m + 1)), x] + Dist[1/(a*c*
g^n*(m + 1)), Int[(g*x)^(m + n)*(a + b*x^n)^p*(c + d*x^n)^q*Simp[a*f*c*(m + 1) - e*(b*c + a*d)*(m + n + 1) - e
*n*(b*c*p + a*d*q) - b*e*d*(m + n*(p + q + 2) + 1)*x^n, x], x], x] /; FreeQ[{a, b, c, d, e, f, g, p, q}, x] &&
IGtQ[n, 0] && LtQ[m, -1]

Rubi steps

\begin {align*} \int \frac {\left (-1+x^4\right )^{3/4} \left (4+x^4\right )}{x^8 \left (-4+x^4\right )} \, dx &=\frac {\left (-1+x^4\right )^{3/4}}{7 x^7}-\frac {1}{28} \int \frac {8-44 x^4}{x^4 \left (-4+x^4\right ) \sqrt [4]{-1+x^4}} \, dx\\ &=\frac {\left (-1+x^4\right )^{3/4}}{7 x^7}+\frac {\left (-1+x^4\right )^{3/4}}{42 x^3}+\frac {1}{336} \int \frac {504}{\left (-4+x^4\right ) \sqrt [4]{-1+x^4}} \, dx\\ &=\frac {\left (-1+x^4\right )^{3/4}}{7 x^7}+\frac {\left (-1+x^4\right )^{3/4}}{42 x^3}+\frac {3}{2} \int \frac {1}{\left (-4+x^4\right ) \sqrt [4]{-1+x^4}} \, dx\\ &=\frac {\left (-1+x^4\right )^{3/4}}{7 x^7}+\frac {\left (-1+x^4\right )^{3/4}}{42 x^3}+\frac {3}{2} \operatorname {Subst}\left (\int \frac {1}{-4+3 x^4} \, dx,x,\frac {x}{\sqrt [4]{-1+x^4}}\right )\\ &=\frac {\left (-1+x^4\right )^{3/4}}{7 x^7}+\frac {\left (-1+x^4\right )^{3/4}}{42 x^3}-\frac {3}{8} \operatorname {Subst}\left (\int \frac {1}{2-\sqrt {3} x^2} \, dx,x,\frac {x}{\sqrt [4]{-1+x^4}}\right )-\frac {3}{8} \operatorname {Subst}\left (\int \frac {1}{2+\sqrt {3} x^2} \, dx,x,\frac {x}{\sqrt [4]{-1+x^4}}\right )\\ &=\frac {\left (-1+x^4\right )^{3/4}}{7 x^7}+\frac {\left (-1+x^4\right )^{3/4}}{42 x^3}-\frac {3^{3/4} \tan ^{-1}\left (\frac {\sqrt [4]{3} x}{\sqrt {2} \sqrt [4]{-1+x^4}}\right )}{8 \sqrt {2}}-\frac {3^{3/4} \tanh ^{-1}\left (\frac {\sqrt [4]{3} x}{\sqrt {2} \sqrt [4]{-1+x^4}}\right )}{8 \sqrt {2}}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]  time = 0.26, size = 118, normalized size = 1.26 \begin {gather*} \frac {\left (x^4-1\right )^{3/4} \left (x^4+6\right )}{42 x^7}-\frac {3^{3/4} \left (-\log \left (2-\frac {\sqrt {2} \sqrt [4]{3} x}{\sqrt [4]{1-x^4}}\right )+\log \left (\frac {\sqrt {2} \sqrt [4]{3} x}{\sqrt [4]{1-x^4}}+2\right )+2 \tan ^{-1}\left (\frac {\sqrt [4]{3} x}{\sqrt {2} \sqrt [4]{1-x^4}}\right )\right )}{16 \sqrt {2}} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

Integrate[((-1 + x^4)^(3/4)*(4 + x^4))/(x^8*(-4 + x^4)),x]

[Out]

((-1 + x^4)^(3/4)*(6 + x^4))/(42*x^7) - (3^(3/4)*(2*ArcTan[(3^(1/4)*x)/(Sqrt[2]*(1 - x^4)^(1/4))] - Log[2 - (S
qrt[2]*3^(1/4)*x)/(1 - x^4)^(1/4)] + Log[2 + (Sqrt[2]*3^(1/4)*x)/(1 - x^4)^(1/4)]))/(16*Sqrt[2])

________________________________________________________________________________________

IntegrateAlgebraic [A]  time = 0.48, size = 94, normalized size = 1.00 \begin {gather*} \frac {\left (-1+x^4\right )^{3/4} \left (6+x^4\right )}{42 x^7}-\frac {3^{3/4} \tan ^{-1}\left (\frac {\sqrt [4]{3} x}{\sqrt {2} \sqrt [4]{-1+x^4}}\right )}{8 \sqrt {2}}-\frac {3^{3/4} \tanh ^{-1}\left (\frac {\sqrt [4]{3} x}{\sqrt {2} \sqrt [4]{-1+x^4}}\right )}{8 \sqrt {2}} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[((-1 + x^4)^(3/4)*(4 + x^4))/(x^8*(-4 + x^4)),x]

[Out]

((-1 + x^4)^(3/4)*(6 + x^4))/(42*x^7) - (3^(3/4)*ArcTan[(3^(1/4)*x)/(Sqrt[2]*(-1 + x^4)^(1/4))])/(8*Sqrt[2]) -
(3^(3/4)*ArcTanh[(3^(1/4)*x)/(Sqrt[2]*(-1 + x^4)^(1/4))])/(8*Sqrt[2])

________________________________________________________________________________________

fricas [B]  time = 6.46, size = 277, normalized size = 2.95 \begin {gather*} \frac {84 \cdot 27^{\frac {1}{4}} \sqrt {2} x^{7} \arctan \left (-\frac {108 \cdot 27^{\frac {1}{4}} \sqrt {2} {\left (x^{4} - 1\right )}^{\frac {1}{4}} x^{3} + 24 \cdot 27^{\frac {3}{4}} \sqrt {2} {\left (x^{4} - 1\right )}^{\frac {3}{4}} x - \sqrt {6} 3^{\frac {1}{4}} {\left (36 \cdot 27^{\frac {1}{4}} \sqrt {2} \sqrt {x^{4} - 1} x^{2} + 27^{\frac {3}{4}} \sqrt {2} {\left (7 \, x^{4} - 4\right )}\right )}}{54 \, {\left (x^{4} - 4\right )}}\right ) - 21 \cdot 27^{\frac {1}{4}} \sqrt {2} x^{7} \log \left (\frac {2 \, {\left (4 \cdot 27^{\frac {3}{4}} \sqrt {2} \sqrt {x^{4} - 1} x^{2} + 36 \, \sqrt {3} {\left (x^{4} - 1\right )}^{\frac {1}{4}} x^{3} + 3 \cdot 27^{\frac {1}{4}} \sqrt {2} {\left (7 \, x^{4} - 4\right )} + 72 \, {\left (x^{4} - 1\right )}^{\frac {3}{4}} x\right )}}{x^{4} - 4}\right ) + 21 \cdot 27^{\frac {1}{4}} \sqrt {2} x^{7} \log \left (-\frac {2 \, {\left (4 \cdot 27^{\frac {3}{4}} \sqrt {2} \sqrt {x^{4} - 1} x^{2} - 36 \, \sqrt {3} {\left (x^{4} - 1\right )}^{\frac {1}{4}} x^{3} + 3 \cdot 27^{\frac {1}{4}} \sqrt {2} {\left (7 \, x^{4} - 4\right )} - 72 \, {\left (x^{4} - 1\right )}^{\frac {3}{4}} x\right )}}{x^{4} - 4}\right ) + 32 \, {\left (x^{4} + 6\right )} {\left (x^{4} - 1\right )}^{\frac {3}{4}}}{1344 \, x^{7}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x^4-1)^(3/4)*(x^4+4)/x^8/(x^4-4),x, algorithm="fricas")

[Out]

1/1344*(84*27^(1/4)*sqrt(2)*x^7*arctan(-1/54*(108*27^(1/4)*sqrt(2)*(x^4 - 1)^(1/4)*x^3 + 24*27^(3/4)*sqrt(2)*(
x^4 - 1)^(3/4)*x - sqrt(6)*3^(1/4)*(36*27^(1/4)*sqrt(2)*sqrt(x^4 - 1)*x^2 + 27^(3/4)*sqrt(2)*(7*x^4 - 4)))/(x^
4 - 4)) - 21*27^(1/4)*sqrt(2)*x^7*log(2*(4*27^(3/4)*sqrt(2)*sqrt(x^4 - 1)*x^2 + 36*sqrt(3)*(x^4 - 1)^(1/4)*x^3
+ 3*27^(1/4)*sqrt(2)*(7*x^4 - 4) + 72*(x^4 - 1)^(3/4)*x)/(x^4 - 4)) + 21*27^(1/4)*sqrt(2)*x^7*log(-2*(4*27^(3
/4)*sqrt(2)*sqrt(x^4 - 1)*x^2 - 36*sqrt(3)*(x^4 - 1)^(1/4)*x^3 + 3*27^(1/4)*sqrt(2)*(7*x^4 - 4) - 72*(x^4 - 1)
^(3/4)*x)/(x^4 - 4)) + 32*(x^4 + 6)*(x^4 - 1)^(3/4))/x^7

________________________________________________________________________________________

giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {{\left (x^{4} + 4\right )} {\left (x^{4} - 1\right )}^{\frac {3}{4}}}{{\left (x^{4} - 4\right )} x^{8}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x^4-1)^(3/4)*(x^4+4)/x^8/(x^4-4),x, algorithm="giac")

[Out]

integrate((x^4 + 4)*(x^4 - 1)^(3/4)/((x^4 - 4)*x^8), x)

________________________________________________________________________________________

maple [C]  time = 11.23, size = 241, normalized size = 2.56

method	result	size
trager	\(\frac {\left (x^{4}-1\right )^{\frac {3}{4}} \left (x^{4}+6\right )}{42 x^{7}}+\frac {\RootOf \left (\textit {\_Z}^{2}+\RootOf \left (\textit {\_Z}^{4}-108\right )^{2}\right ) \ln \left (\frac {2 \RootOf \left (\textit {\_Z}^{2}+\RootOf \left (\textit {\_Z}^{4}-108\right )^{2}\right ) \sqrt {x^{4}-1}\, \RootOf \left (\textit {\_Z}^{4}-108\right )^{2} x^{2}-6 \left (x^{4}-1\right )^{\frac {1}{4}} \RootOf \left (\textit {\_Z}^{4}-108\right )^{2} x^{3}-21 \RootOf \left (\textit {\_Z}^{2}+\RootOf \left (\textit {\_Z}^{4}-108\right )^{2}\right ) x^{4}+72 \left (x^{4}-1\right )^{\frac {3}{4}} x +12 \RootOf \left (\textit {\_Z}^{2}+\RootOf \left (\textit {\_Z}^{4}-108\right )^{2}\right )}{\left (x^{2}-2\right ) \left (x^{2}+2\right )}\right )}{32}-\frac {\RootOf \left (\textit {\_Z}^{4}-108\right ) \ln \left (\frac {2 \sqrt {x^{4}-1}\, \RootOf \left (\textit {\_Z}^{4}-108\right )^{3} x^{2}+6 \left (x^{4}-1\right )^{\frac {1}{4}} \RootOf \left (\textit {\_Z}^{4}-108\right )^{2} x^{3}+21 \RootOf \left (\textit {\_Z}^{4}-108\right ) x^{4}+72 \left (x^{4}-1\right )^{\frac {3}{4}} x -12 \RootOf \left (\textit {\_Z}^{4}-108\right )}{\left (x^{2}-2\right ) \left (x^{2}+2\right )}\right )}{32}\)	\(241\)
risch	\(\frac {x^{8}+5 x^{4}-6}{42 x^{7} \left (x^{4}-1\right )^{\frac {1}{4}}}+\frac {\RootOf \left (\textit {\_Z}^{2}+\RootOf \left (\textit {\_Z}^{4}-108\right )^{2}\right ) \ln \left (\frac {2 \RootOf \left (\textit {\_Z}^{2}+\RootOf \left (\textit {\_Z}^{4}-108\right )^{2}\right ) \sqrt {x^{4}-1}\, \RootOf \left (\textit {\_Z}^{4}-108\right )^{2} x^{2}-6 \left (x^{4}-1\right )^{\frac {1}{4}} \RootOf \left (\textit {\_Z}^{4}-108\right )^{2} x^{3}-21 \RootOf \left (\textit {\_Z}^{2}+\RootOf \left (\textit {\_Z}^{4}-108\right )^{2}\right ) x^{4}+72 \left (x^{4}-1\right )^{\frac {3}{4}} x +12 \RootOf \left (\textit {\_Z}^{2}+\RootOf \left (\textit {\_Z}^{4}-108\right )^{2}\right )}{\left (x^{2}-2\right ) \left (x^{2}+2\right )}\right )}{32}-\frac {\RootOf \left (\textit {\_Z}^{4}-108\right ) \ln \left (\frac {2 \sqrt {x^{4}-1}\, \RootOf \left (\textit {\_Z}^{4}-108\right )^{3} x^{2}+6 \left (x^{4}-1\right )^{\frac {1}{4}} \RootOf \left (\textit {\_Z}^{4}-108\right )^{2} x^{3}+21 \RootOf \left (\textit {\_Z}^{4}-108\right ) x^{4}+72 \left (x^{4}-1\right )^{\frac {3}{4}} x -12 \RootOf \left (\textit {\_Z}^{4}-108\right )}{\left (x^{2}-2\right ) \left (x^{2}+2\right )}\right )}{32}\)	\(246\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^4-1)^(3/4)*(x^4+4)/x^8/(x^4-4),x,method=_RETURNVERBOSE)

[Out]

1/42*(x^4-1)^(3/4)*(x^4+6)/x^7+1/32*RootOf(_Z^2+RootOf(_Z^4-108)^2)*ln((2*RootOf(_Z^2+RootOf(_Z^4-108)^2)*(x^4
-1)^(1/2)*RootOf(_Z^4-108)^2*x^2-6*(x^4-1)^(1/4)*RootOf(_Z^4-108)^2*x^3-21*RootOf(_Z^2+RootOf(_Z^4-108)^2)*x^4
+72*(x^4-1)^(3/4)*x+12*RootOf(_Z^2+RootOf(_Z^4-108)^2))/(x^2-2)/(x^2+2))-1/32*RootOf(_Z^4-108)*ln((2*(x^4-1)^(
1/2)*RootOf(_Z^4-108)^3*x^2+6*(x^4-1)^(1/4)*RootOf(_Z^4-108)^2*x^3+21*RootOf(_Z^4-108)*x^4+72*(x^4-1)^(3/4)*x-
12*RootOf(_Z^4-108))/(x^2-2)/(x^2+2))

________________________________________________________________________________________

maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {{\left (x^{4} + 4\right )} {\left (x^{4} - 1\right )}^{\frac {3}{4}}}{{\left (x^{4} - 4\right )} x^{8}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x^4-1)^(3/4)*(x^4+4)/x^8/(x^4-4),x, algorithm="maxima")

[Out]

integrate((x^4 + 4)*(x^4 - 1)^(3/4)/((x^4 - 4)*x^8), x)

________________________________________________________________________________________

mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {{\left (x^4-1\right )}^{3/4}\,\left (x^4+4\right )}{x^8\,\left (x^4-4\right )} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((x^4 - 1)^(3/4)*(x^4 + 4))/(x^8*(x^4 - 4)),x)

[Out]

int(((x^4 - 1)^(3/4)*(x^4 + 4))/(x^8*(x^4 - 4)), x)

________________________________________________________________________________________

sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (\left (x - 1\right ) \left (x + 1\right ) \left (x^{2} + 1\right )\right )^{\frac {3}{4}} \left (x^{2} - 2 x + 2\right ) \left (x^{2} + 2 x + 2\right )}{x^{8} \left (x^{2} - 2\right ) \left (x^{2} + 2\right )}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x**4-1)**(3/4)*(x**4+4)/x**8/(x**4-4),x)

[Out]

Integral(((x - 1)*(x + 1)*(x**2 + 1))**(3/4)*(x**2 - 2*x + 2)*(x**2 + 2*x + 2)/(x**8*(x**2 - 2)*(x**2 + 2)), x
)

________________________________________________________________________________________